2007 AMC 12A Problem 10

Attempt Problem 10 of the 2007 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2007 AMC 12A solutions, or check the answer key.

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10.

A triangle with side lengths in the ratio

3:4:5

is inscribed in a circle of radius

3.

What is the area of the triangle?

8.64

12

5\pi

17.28

18

3\!:\!4\!:\!5

triangle is right-angled, so its hypotenuse is a diameter of the circle

The hypotenuse

5x

equals the diameter

6,

then the area is

\tfrac12(3x)(4x)

Let the sides be

3x,

4x,

and

5x.

The triangle is right, so its hypotenuse is a diameter.

Thus

5x=2\cdot 3=6,

giving

x=\tfrac65.

The area is

\tfrac12\cdot 3x\cdot 4x=6x^2

=6\cdot\tfrac{36}{25}

=\tfrac{216}{25}=8.64.

Thus, the correct answer is A.

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